HPLC Instrumental Analysis PDF

Department of l Pharmaceutica a l y t i c a l C he m istry An  h p e r f o r m a n c e l i q u i d H i g o m a t o g r a p h y ( H P LC ) Ch r  Definition of terms related to the chromatogram Solvent front Base line term...

Department of l Pharmaceutica a l y t i c a l C he m istry An  h p e r f o r m a n c e l i q u i d H i g o m a t o g r a p h y ( H P LC ) Ch r  Definition of terms related to the chromatogram Solvent front Base line term Definition That portion of the chromatogram recorded when only the mobile phase Base Line emerges from the column. That portion of the chromatogram recorded while a single component emerges Peak from the column (or, if separation of a mixture has been incomplete, two or more unseparated components may appear as one peak).  Peak area Peak height Peak base Peak width Peak Base An interpolation of the base line between the extremities of the peak. Peak Area The area enclosed by the peak and the peak base Peak Height The distance from the peak maximum to the peak base. The magnitude of the peak base intercepted by the tangents to the Peak Width (W) inflection points of the peak.  The time between sample injection and the appearance of a Retention time (tR) solute peak at the detector of a chromatographic column. The time it takes for an unretained species (mobile phase) to pass Dead time (tM) through a column. Adjusted retention time (tR) is the time solute spent in the stationary phase, and equals to [tR-tM].  The effectiveness of chromatographic column in separating two solutes depends on several factors. ❖The capacity factor (K) is widely used to describe the migration rates of solutes on columns (measuring the retention of the solute on the column). It is defined as: 𝒕𝑹 𝑨−𝒕𝑴 𝒕𝑹 𝑩−𝒕𝑴 for solute A : KA = 𝒕𝑴 and KB = 𝒕𝑴 for solute B. ❖Ideally, separations are performed under conditions in which the capacity factors For the solutes in a mixture lie in the range between 1 and 5. ❖The selectivity factors (α) of a column for the two solutes a and b in a mixture is the ability of the chromatographic system to distinguish between sample content Defined as: Where KB is the capacity factor of the more retained solute B, and KA is that for the More rapidly eluted solute A. By this definition, α is always greater than unity. The larger its value, the more selective the stationary phase and the easier the separation will be.  Efficiency of chromatographic columns ❖The efficiency of chromatographic column refers to the amount of band broading that occurs when a compound passes through the column. ❖Two related terms are widely used as quantitative measures of chromatographic column efficiency. ❖Plate height,(H), (is also known as the height equivalent to theoretical plate (HETP), and number of theoretical plates (N). ❖The two are related by the equation : N = L / h (Where L is the column length, usually in centimeters) ❖The efficiency of chromatographic columns increases as the number of plates Becomes greater and as the plate height becomes smaller. ❖The two parameters N and H can be calculated using the equations. ❖The resolution (Rs) of a column provides a quantitative measure of its ability to separate two analytes. ❖The resolution of each column is defined as:  Interpretation of Chromatograms The data acquired from the chromatogram: tM, tRA, tRB, WA, WB, hA, hB, …………… cm The required chromatographic parameters:  E.G. For chromatograms 1- chromatographic determination of a two components mixture Where: (a) solvent front (b), (c) are the two components ❖Stationary phase: C8 column (length = 25 cm) ❖Mobile phase: acetonitrile : methanol : 0.02 M phosphate buffer (ph 5) [20:20:60] ❖Detector: uv-detector at 226 nm ❖Injection volume: 20 µl ❖Flow rate : 1 ml/ min Parameter Value 1- NTP 2- HETP 3- K' for compounds b and c CONSIDERING PREVIOUS FIGURE, COMPLETE THE FOLLOWING TABLE: 4- Resolution 5- Selectivity factor  solvent Samle b Sample c Retention time tM= 1.82 min. (tR)b =3.01 min. (tR)c = 5.3 min. Peak width (w)B = 0.4 cm (w)C = 0.6 cm Parameter law Value For B: 𝟑.𝟎𝟏 𝑡𝑅 NTP=16( 𝟎.𝟒 )2 = 906.01 1- NTP NTP=16( 𝑤 )2 For C: 𝟓.𝟑 NTP=16(𝟎.𝟔)2 = 1248.44 For B: 𝟐𝟓 𝐿 HETP = 𝟗𝟎𝟔.𝟎𝟏 = 0.028 cm 2- HETP HETP= 𝑁𝑇𝑃 For C: 𝟐𝟓 HETP = 𝟏𝟐𝟒𝟖.𝟒𝟒= 0.02 cm  Answer : For B: 𝟑.𝟎𝟏−𝟏.𝟖𝟐 𝑡𝑅−𝑡𝑀 K' = 𝟏.𝟖𝟐 = 0.65 3- K' for compounds b and c K' = 𝑡𝑀 For C: 𝟓.𝟑−𝟏.𝟖𝟐 K' = 𝟏.𝟖𝟐 = 1.91 (𝑡𝑅)𝑐−(𝑡𝑅)𝑏 𝟓.𝟑−𝟑.𝟎𝟏 4- Resolution Rs= 2( ) Rs= 2( ) = 4.58 𝑤 𝑏+ 𝑤 𝑐 𝟎.𝟒 +𝟎.𝟔 𝑡𝑅 𝑐−𝑡𝑀 𝟓.𝟑 −𝟏.𝟖𝟐 5- Selectivity factor α= α= 𝟑.𝟎𝟏−𝟏.𝟖𝟐 = 2.92 𝑡𝑅 𝑏−𝑡𝑀